Question

Let a curve y = y(x) pass through the point (3, 3) and the area of the region under this curve, above the x-axis and between the abscissae 3 and 
\(x(>3)\ be\ (\frac{y}{x})^3\)
. If this curve also passes through the point (α,6√10) in the first quadrant, then α is equal to _______.

Answer 1

Correct Answer: 6

\(\int_{3}^{x} \, f(x) \,dx = \left(\frac{f(x)}{x}\right)^3\)
\(x^3 \cdot \int_{3}^{x} f(x) \,dx = f^3(x)\)
Differentiate w.r.t. x
\(x^3 f(x) +3x^2. \frac{f^3(x)}{x^3} = 3f^2(x)f'(x)\)
\(⇒\)\(3y^2 \frac{dy}{dx} = x^3y + \frac{3y^3}{x}\)
\(3xy \frac{dy}{dx}=x^4+3y^2\)
Let y² = t
\(\frac{3}{2} \frac{dt}{dx} = x^3 + \frac{3t}{x}\)
\(\frac{dt}{dx} - \frac{2t}{x} = \frac{2x^3}{3}\)
\(IF=e^{\int−\frac{2}{x} dx}=\frac{1}{x^2}\)
Solution of differential equation
\(t⋅\frac{1}{x^2}=\int\frac{2}{3}x\ dx\)
\(\frac{y^2}{x^2}=\frac{x^2}{3}+C\)
\(y^2=\frac{x^4}{3}+Cx^2\)
Curve passes through(3,3)
\(⇒C=–2\)
\(y^2=\frac{x^4}{3}−2x^2\)
Which passes through\( (α, 6\sqrt10)\)
\(\frac{α^4−6α^2}{3}=360\)
\(α^4−6α^2−1080=0\)
α=6
So, the correct answer is 6.